What is the area of the region between the graphs of $f(x)=8x+6$ and $g(x)=x-x^2$ from $x=-6$ to $x=-1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{485}{6}$ (Choice B) B $\dfrac{1195}{6}$ (Choice C) C $\dfrac{643}{6}$ (Choice D) D $\dfrac{125}{6}$
Visualizing the area We sketch the graphs of $f$ and $g$ first. ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}10}$ ${\llap{-}20}$ ${\llap{-}30}$ ${\llap{-}40}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between $x=-6$ and $x=-1$. From this we are looking to evaluate: $ \int_{-6}^{-1}\left( g(x)-f(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{-6}^{-1} \left( x-x^2- (8x+6) \right) \,dx \\\\ &= \int_{-6}^{-1} \left( -x^2- 7x-6 \right) \,dx\\\\ &= -\dfrac{x^3}{3}-\dfrac{7x^2}{2}-6x~\Bigg|_{-6}^{-1} \\\\ &= \left( \dfrac{1}{3}-\dfrac{7}{2}+6 \right) -\left( 72-126+36 \right)\\\\ &=\dfrac{125}{6} \end{aligned}$ Answer The area is $\dfrac{125}{6}$ square units.